LintCode 236: Swap Bits (位運算經典題)
阿新 • • 發佈:2018-12-19
我之前的做法太繁瑣。
class Solution { public: /* * @param x: An integer * @return: An integer */ int swapOddEvenBits(int x) { int len = sizeof(int) * 8; for (int i = 0; i < len; i += 2) { int even = (x >> i) & 0x1; int odd = (x >> i + 1) & 0x1; if (odd != even) { if (even == 1) { x &= ~(0x1 << i); //clear i th bit x |= (0x1 << (i + 1)); //set the i+1 th bit } else { x &= ~(0x1 << (i + 1)); //clear i+1 th bit x |= (0x1 << i); //set the i th bit } } } return x; } };
發現最簡單的做法就是:
int swapOddEvenBits(int x) {
return ((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1);
}